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We need to use spherical trigonometry, in which the sides of triangles are arcs of great circles, and are measured in the same units as angles. For convenience we'll use degrees rather than radians. The locations of London and Tokyo, in decimal degrees, are 51.500 N, 0.167 W and 35.667 N, 139.750 E respectively. (We'll keep three decimal places for the degrees in the intermediate calculations.)
The formulas of spherical trig that we will use are the law of cosines:
cos a = cos b cos c + sin b sin c cos A, where lower case designates a side and upper case an angle, and the side opposite a given angle has the same letter of the alphabet,
and the law of sines:
sin a / sin A = sin b / sin B.
First find the arc distance (angular measure) from London to Tokyo, by solving a triangle with vertices at London, Tokyo and the North Pole. Remembering that the arc from a location to the north pole is the complement of its latitude, latitudes can be used in the cosine formula in place of the respective sides so long as the sine and cosine function are interchanged. The angle opposite the side that represents the distance is at the North Pole and is the west longitude of London added to the east longitude of Tokyo.
cos (dist) = sin 51.5 sin 35.667 + cos 51.5 cos35.667 cos(139.917)
making the distance 86.022 degrees.
Consider the triangle formed by Tokyo, London and point X, with the Tokyo-London base at the top and point X at the bottom. Call it triangle TLX. The North Pole lies within this triangle as the side LT lies on the other side from the United States. Side LT is 86.022° as we have just calculated. Side LX is the distance from London to point X converted to degrees. This distance is 4386/3959 radians, or 63.475°. Side TX is 6182/3959 radians, or 89.468°.
Angle L of this triangle can be calculated from
Cos 89.468 = cos 86.022 cos 63.475 + sin 86.022 sin 63.475 cos L
L = 91.393°
Specifically that is angle TLX.
We need to make a smaller triangle, with the North Pole, London and point X at its vertices; call it triangle NLX. Once we solve it we'll know where point X is. Remember point N (the North Pole) is within the larger triangle. We already know side NL is 90°-51.5°, and side XL = 63.475°. We can solve the whole triangle if we get one more piece of it. The way we can get that is to take angle TLX and subtract angle TLN. That is in the same triangle we considered in computing the distance from Tokyo to London, with two sides equal to the colatitudes of the respective cities and the angle between them the difference of their longitudes (or adding their absolute values as they had opposite signs). Here we can use the law of sines. sin 86.022 / sin 139.917 = cos 35.667 / sin TLN. Note the cosine of 35.667 is used as that's the sine of the corresponding distance from the pole. Using this, angle TLN comes out to 31.626. Subtracting this from angle TLX gives NLX = 59.767°. Thus we know two sides and the included angle of triangle NLX. We can use the law of cosines to get the third side, or by shifting to the sine, the latitude of point X. sin(lat) = sin 51.5 cos 63.475 + cos 51.5 sin 63.475 cos 59.767. The latitude thus comes out to 39.047°.
The difference in longitude between point X and London is angle XNL, opposite side XL. The law of cosines gives cos 63.475 = sin 39.047 sin 51.5 + cos 39.047 cos 51.5 cos XNL. Solving we get XNL = 95.510°. When this is added to London's longitude of 0.167° we get the longitude of point X as 95.677°.
Point X is therefore at 39.05° N, 95.68° W. This is in Topeka, Kansas.
For more information on finding distances on the real planet earth (oblate rather than perfectly spherical), Guy Bomford's out-of-print book Geodesy has appropriate formulas, particularly Rudoe's formula, complicated, but suited to a computer's speed. |
